An unknown element #"X"# has electron configuration #["Ar"]3d^5# and a charge of #3+#. What is the atomic number of #"X"#?

For starters, you know that your element is located in period #4# of the Periodic Table because all the transition metals that have #3d# configurations are located in period #4# of the Periodic Table.

This means that you can write the noble-gas shorthand configuration for this unknown element, let's say #"X"#, like this

#"X"^(3+): ["Ar"] 3d^5#

Now, the trick here is to realize that when a transition metal located in period #4# is losing electrons, the first two electrons that are lost are coming from the #4s# orbital.

In other words, when the #4s# orbital is occupied, it is actually higher in energy than the #3d# orbitals. Keep in mind that when the #4s# orbital is empty, it is actually lower in energy than the #3d# orbitals!

So when #"X"# is losing the first two electrons, they are being taken from the #4s# orbital. Consequently, you can say that when the element is losing the third electron, this electron will come from the #3d# orbitals.

You can thus backtrack and add electrons to the #3+# cations to form the neutral atom.

#"X"^(2+) = "X"^(3+) + "e"^(-)#

This means that you have--remember, the third electron was removed from the #3d# orbitals, so it must be added back to the #3d# orbitals!

#"X"^(2+): ["Ar"] 3d^6#

Next, you have

#"X" = "X"^(2+) + 2"e"^(-)#

This means that you have--the first two electrons were removed from the #4s# orbital, so they must be added back to the #4s# orbital!

#"X": ["Ar"]3d^6 4s^2#

Therefore, you can say that a neutral atom of #"X"# has a total of

#overbrace("18 e"^(-))^(color(blue)("the same as neutral atom of Ar")) + overbrace("6 e"^(-))^(color(blue)("in the 3d orbitals")) + overbrace("2 e"^(-))^(color(blue)("in the 4s orbitals")) = "26 e"^(-)#

This implies that a neutral atom of #"X"# must have #26# protons located inside its nucleus.

You can conclude that your unknown element is iron, #"Fe"#, which has an atomic number equal to #26#.